# Conditional probability for continuous function

let $f$ is a continuous probability density function. Then a conditional probability is $f\left( x|Y=y \right)=\frac{{{f}_{XY}}\left( x,y \right)}{{{f}_{Y}}\left( y \right)}$. For example if we want $Y=1$ we put this in ${{f}_{Y}}\left( y \right)$. But this is $0$ because $f$ is a continuous function and denominator is $0$. so how can we interpret conditional distribution for continuous function? ... As a limit?

It isn't 0. If $f_Y(y)=0 \forall y$ then things would be a lot different. Consider for example an exponential density: $$f_Y(y)=\frac{1}{\theta}e^{-\frac{y}{\theta}}$$

For $y$ in the set $\{y \in \mathbb{R} : y>0\}$ then we have that $f_Y(y)>0$. So for instance using your example of $y=1$: $$f_Y(1)=\frac{1}{\theta}e^{-\frac{1}{\theta}}>0$$ The distinction is that this does not describe a probability. The probability for this distribution is given by some interval (and the area under the graph), and not by a single point. So no, a conditional distribution is not a limit.

This is not very rigorous, but I think it's helpful for the intuition. The probability is approximately the height of the density times the interval. So for example in a $U(0,1)$ distribution, the height is 1 because $f_Y(y)=1$, $0\leq y\leq 1$. And then for the interval $[0.25, 0.50]$ the probability is $(0.50-0.25)*1=0.25$. This is then the integral:

$$\int_{0.25}^{0.50}1dy=0.25$$

Consider this but in more general form:

$$\int_{a}^{b}1dy$$

When we look at a smaller and smaller interval, what happens? Well:

$$\lim_{b\to a}\int_{a}^b1dy=a-a=0$$

As the interval gets smaller, the probability goes to 0. Hence, a single point has probability 0.

Consider now a discrete distribution. Here we cannot look at such small intervals. The smallest (non-zero) interval is 1. For example, a discrete uniform on $[1, 10]$:

$$p_X(x)=\frac{1}{10}, \quad x=1, \dots, 10$$ The height is $1/10$, and now we can think of $x=2$ as sort of the interval from 2 to 3. Its length is 1 so height times interval yields $1/10*1=1/10$. So I think that it can help the intuition to think of the discrete case as intervals of length 1, which is the smallest interval we can have. In the continuous case we can pick smaller and smaller intervals eventually giving a zero probability, but we cannot here.